Problem: Find the unit vector in the direction of $\vec{v}=\left( 6, 3 \right)$. $($ $~,$ $)$
Explanation: Getting started A unit vector has a magnitude (or length) of $1$. Dividing $\vec v$ by its magnitude will find a vector in the same direction as $\vec v$ but with a magnitude of $1$ : $\text{Unit vector in the direction of } \vec v = \dfrac{\vec v}{|| \vec v||}$ Finding the unit vector $\begin{aligned} \dfrac{\vec{v}}{||\vec{v}||} &= \dfrac{\left( 6, 3 \right) }{\sqrt{6^2+3^2}} \\\\\\ &= \dfrac{1}{\sqrt{45}} \cdot \left( 6, 3 \right) \\\\\\ &= \left( {\dfrac{6}{\sqrt{45}}}, {\dfrac{3}{\sqrt{45}}}\right) \end{aligned}$ Great, we found the unit vector! Just to be careful, let's check and make sure it has a magnitude of $1$. Verifying that the magnitude is $1$ $\begin{aligned}&\sqrt{\left( {\dfrac{6}{\sqrt{45}}} \right)^2 + \left( {\dfrac{3}{\sqrt{45}}} \right)^2} \\\\\\ &= \sqrt{\dfrac{36}{45} + \dfrac{9}{45}} \\\\\\ &= \sqrt{\dfrac{45}{45}} \\\\\\ &=1 \end{aligned}$ The answer $\left( {\dfrac{6}{\sqrt{45}}}, {\dfrac{3}{\sqrt{45}}}\right) $ Visualizing the answer: $6$ $3$ $\dfrac{6}{\sqrt{45}}$ $\dfrac{3}{\sqrt{45}}$ $\vec{v}$ ${\text{unit vector}}$ Notice how the unit vector points in the same direction as the original vector, but the unit vector has a magnitude of $1$.